Finding the oblique asymptote of $f(x)=2x\tan^{-1}(x)$ I have problems with the computation of the oblique asymptote for the function
$$f(x)=2x\tan^{-1}(x)$$
I started by stating that $y = mx + b$ is the equation of the oblique asymptote. Next I looked at the following limit
$$\lim_{x \rightarrow \infty } \left(2x \tan^{-1}(x) - mx - b\right) = 0 $$
from where I know that since $m$ and $b$ are constants, that I must have: 
$$\lim_{x \rightarrow \infty }\left(2x\tan^{-1}(x) - mx\right) = b$$
thus we must have $$\lim_{x \rightarrow \infty }\left(2x\tan^{-1}(x) - mx\right) = 0$$
and hence $$x\lim_{x \rightarrow \infty }\left(2\tan^{-1}(x) - m\right) = 0$$
which does not exists unless we have that $m=\pi$. From here I do not know how to proceed. I tried plugging in my value for $m$ - but without luck. 
I found a similar question here, Oblique asymptotes of $f(x)=\frac{x}{\arctan x}$ , and tried to follow it, but without luck. 
Extra: 

Best regards 
 A: Let $y=mx+b$ be the oblique asymptote as $x\rightarrow\infty$.
Then $\displaystyle\lim_{x\to\infty}\left(2x\tan^{-1}(x)-mx-b\right)=0$, 
so $\displaystyle\lim_{x\to\infty}\left(2x\tan^{-1}(x)-mx\right)=b$  
We have that $\displaystyle\lim_{x\to\infty}\left(2x\tan^{-1}(x)-mx\right)=\lim_{x\to\infty}\left(2\tan^{-1}(x)-m\right)x$ doesn't exist if $\displaystyle{\lim_{x\to\infty}\left(2\tan^{-1}(x)-m\right)\neq 0 \Rightarrow \lim_{x\to\infty}\tan^{-1}(x)\neq \frac{m}{2}\Rightarrow \frac{\pi}{2}\neq \frac{m}{2} \Rightarrow m\neq \pi}$. 
So, $\displaystyle m=\pi$, and then $\displaystyle\lim_{x\to\infty}\left(2x\tan^{-1}(x)-\pi x\right)=\lim_{x\rightarrow  0}\left(2\frac{1}{x}\tan^{-1}\left (\frac{1}{x}\right )-\pi \frac{1}{x}\right )=\lim_{x\rightarrow  0}\left(\frac{2\tan^{-1}\left (\frac{1}{x}\right )-\pi }{x}\right ) \ \overset{\text{ De l'Hospital }}{ = } \ \lim_{x\rightarrow  0}\left(2\frac{-1}{x^2+1}\right )=-2$ 
so $b=-2$. 
Therefore, $\displaystyle y=\pi x-2$ is the oblique asymptote. 
A: $ax+b$ being an asymptote to $2x\tan^{-1}x$ is equivalent to the statement;
$$2x\tan^{-1}x-ax-b=o(1)$$
as $x\to\infty$. Now, dividing by $x$ tells us
$$2\tan^{-1}x-a=o(1)\implies a=\lim_{x\to\infty}2\tan^{-1}x=\pi$$
This leads us to a refinement of our first statement, seeing that
$$2x\tan^{-1}x-\pi x-b=o(1)\implies b=\lim_{x\to\infty}2x\tan^{-1}x-\pi x$$
We now split this expression as $2x\left(\tan^{-1}x-\frac{\pi}{2}\right)$, and note that for $x>0$, $\tan^{-1}x=\frac{\pi}{2}-\tan^{-1}\left(\frac{1}{x}\right)$, whence:
$$b=\lim_{x\to\infty}\left(-2x\tan^{-1}\left(\frac{1}{x}\right)\right)$$
Let $x=\frac{1}{y}$ to reframe this limit as $\lim_{y\to0}\frac{-2\tan^{-1}y}{y}$ and use L'Hopital's Rule / the definition of the derivative / any of a number of other methods to retrieve this limit as $-2$.
We are thus assured that:
$$2x\tan^{-1}x=\pi x-2+o(1)$$
i.e. that the asymptote as $x\to\infty$ of $2x\tan^{-1}x$ is $\pi x-2$.
A: Use an asymptotic expansion.
First set $x=\dfrac1u$ and remember $\;\arctan x=\begin{cases}\dfrac\pi2-\arctan  u &\text{if}\enspace x>0,\\
-\dfrac\pi2-\arctan u &\text{if}\enspace x<0.
\end{cases}$ 
Now $\arctan u=u+o(u)$, whence, if $x>0$,
$$2x\arctan x=\begin{cases}\pi x-2+o\Bigl(\dfrac1x\Bigr)&\text{if}\enspace x>0,\\
-\pi x-2+o\Bigl(\dfrac1x\Bigr) &\text{if}\enspace x<0.
\end{cases}$$
This proves the oblique asymptotes are
\begin{align*}
y&=\pi x-2\quad\text{when}\quad x\to+\infty\\
y&=-\pi x-2\quad\text{when}\quad x\to-\infty
\end{align*}
Furthermore, if one uses an expansion at order $3$ for $\arctan u$, one can check the curve is above each of its asymptotes near $\infty$.
